If x=k satisfies the equation left|begin{arra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the determinant equation: $\left|\begin{array}{ccc}x-2 & 3(x-1) & 5(x-1) \ x-4 & 3(x-3) & 5(x-5) \ x-8 & 3(x-9) & 5(x-25)\end{array}ight

Vedclass · Accepted Answer

$x^2+2x-3=0$

Vedclass · Answer

(D) Given the determinant equation: $\left|\begin{array}{ccc}x-2 & 3(x-1) & 5(x-1) \ x-4 & 3(x-3) & 5(x-5) \ x-8 & 3(x-9) & 5(x-25)\end{array}ight|=0$. Taking common factors $3$ and $5$ from $C_2$ and $C_3$ respectively: $15 \left|\begin{array}{ccc}x-2 & x-1 & x-1 \ x-4 & x-3 & x-5 \ x-8 & x-9 & x-25\end{array}ight|=0$. Applying $C_2 ightarrow C_2 - C_3$: $\left|\begin{array}{ccc}x-2 & 0 & x-1 \ x-4 & 2 & x-5 \ x-8 & 16 & x-25\end{array}ight|=0$. Expanding along $C_2$: $-2((x-2)(x-25) - (x-1)(x-8)) + 16((x-2)(x-5) - (x-1)(x-4)) = 0$. $-2(x^2-27x+50 - (x^2-9x+8)) + 16(x^2-7x+10 - (x^2-5x+4)) = 0$. $-2(-18x+42) + 16(-2x+6) = 0$. $36x - 84 - 32x + 96 = 0 \Rightarrow 4x + 12 = 0 \Rightarrow x = -3$. Checking the options for $x = -3$: For $x^2+2x-3=0$, $(-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0$. Thus, $x = -3$ satisfies $x^2+2x-3=0$.

If $x=k$ satisfies the equation $\left|\begin{array}{ccc}x-2 & 3x-3 & 5x-5 \\ x-4 & 3x-9 & 5x-25 \\ x-8 & 3x-27 & 5x-125\end{array}\right|=0$, then $x=k$ also satisfies the equation

Similar Questions

The value of $a$ for which the matrix $A = \begin{bmatrix} a & 2 \\ 2 & 4 \end{bmatrix}$ is singular is:

Let $A=\begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3A)))=2^m \cdot 3^n$,where $m, n \in N$,then $m+n$ is equal to:

The determinant $\left| \begin{array}{ccc} a & b & a - b \\ b & c & b - c \\ 2 & 1 & 0 \end{array} \right|$ is equal to zero if $a, b, c$ are in

If the system of equations
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = k+3$
$x + y = 1$
is consistent,then the value of $k$ is

If $(x, y, z) \ne (0, 0, 0)$ and $(i + j + 3k)x + (3i - 3j + k)y + (-4i + 5j)z = \lambda (xi + yj + zk),$ then the value of $\lambda$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module